Re: the number of roots of a polynomial in the first quadrant

Certainly, I need to use the argument principle, but I don't see how. Here's what I tried.

All zeroes of lie in which follows from applying the Cauchy bound. Suppose I can prove that has no real and no imaginary roots. Then, since the function is holomorphic, I have to calculate

where

(integrals have to be calculated in the appropriate directions of course).

But to calculate those integrals, don't I need to know the zeroes of Only then would I be able to integrate by residues, right? It's supposed to be very simple.

For a start, f(z) has no positive real roots (it has the value 1 at z=0, and thereafter it increases, as you can check by elementary calculus). The polynomial has real coefficients, so its non-real roots occur in complex conjugate pairs. Therefore you can look for the number of roots with positive real part, and divide that number by 2 to get the number in the first quadrant.

So rather than looking at the winding number round a quadrant, I would look for the winding number of f(z) round a D-shaped contour consisting of a semicircle in the right-hand half plane, going from –iR to +iR, followed by a straight line segment down the imaginary axis from +iR to -iR. Here, R can be any number greater than or equal to 4.

To work out the winding number of f(z) as z goes round this contour, I would think in purely geometric terms rather than trying to evaluate the integral of f'/f. As z goes round the semicircle, f(z) is dominated by the term , so it goes 4 times round the origin. For the straight line portion of the contour, write z=iy. Then The real part of this is clearly always positive, so f(iy) cannot encircle the origin at all. Thus the winding number remains at 4, and the number of roots in the first quadrant is 2.